Given that frac{1}{3} lambda_m^infty (Fe^{3+} | vedclass

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(A) According to Kohlrausch's law of independent migration of ions,the equivalent conductivity at infinite dilution is the sum of the equivalent condu

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$148$

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(A) According to Kohlrausch's law of independent migration of ions,the equivalent conductivity at infinite dilution is the sum of the equivalent conductivities of the individual ions.$\lambda_{eq}^\infty (Fe_2(SO_4)_3) = \lambda_{eq}^\infty (Fe^{3+}) + \lambda_{eq}^\infty (SO_4^{2-})$Given that $\lambda_{eq}^\infty (Fe^{3+}) = \frac{1}{3} \lambda_m^\infty (Fe^{3+}) = 68 \ \Omega^{-1} \ cm^2 \ eq^{-1}$ and $\lambda_{eq}^\infty (SO_4^{2-}) = \frac{1}{2} \lambda_m^\infty (SO_4^{2-}) = 80 \ \Omega^{-1} \ cm^2 \ eq^{-1}$.Therefore,$\lambda_{eq}^\infty (Fe_2(SO_4)_3) = 68 + 80 = 148 \ \Omega^{-1} \ cm^2 \ eq^{-1}$.

Given that $\frac{1}{3} \lambda_m^\infty (Fe^{3+}) = 68 \ \Omega^{-1} \ cm^{2} \ eq^{-1}$ and $\frac{1}{2} \lambda_m^\infty (SO_4^{2-}) = 80 \ \Omega^{-1} \ cm^{2} \ eq^{-1}$. What will be the value of $\lambda_{eq}^\infty (Fe_2(SO_4)_3)$? ............ $\Omega^{-1} \ cm^2 \ eq^{-1}$

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